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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 3 · Set 2 · Q34
Show that f: R_+ →[-5, ∞) given by f ( x )=4 x^2+4 x -5 is both one-one and onto where R_+=[0, ∞). Also, find p ∈ R_+ such that f ( p )=3.
Show that $\displaystyle \mathrm{f}: \mathrm{R}_{+} \rightarrow[-5, \infty)$ given by $\displaystyle \mathrm{f}(\mathrm{x})=4 \mathrm{x}^{2}+4 \mathrm{x}-5$ is both one-one and onto where $\displaystyle \mathrm{R}_{+}=[0, \infty)$. Also, find $\displaystyle \mathrm{p} \in \mathrm{R}_{+}$such that $\displaystyle \mathrm{f}(\mathrm{p})=3$.
Marking-scheme solution
One$\displaystyle -$One: For $\displaystyle x_{1}, x_{2} \in R_{+}$,
Let $\displaystyle f(x_{1})=f(x_{2}) \Rightarrow 4 x_{1}^{2}+4 x_{1}-5=4 x_{2}^{2}+4 x_{2}-5$
$\displaystyle \Rightarrow (x_{1}-x_{2})(x_{1}+x_{2}+1)=0 \Rightarrow x_{1}=x_{2}$ as $\displaystyle x_{1}+x_{2}+1 \neq 0$ as $\displaystyle x_{1}, x_{2} \in R_{+}$
Hence f is one $\displaystyle -$ one.
Onto: $\displaystyle y=4 x^{2}+4 x-5 \Rightarrow x=\dfrac{\sqrt{y+6}-1}{2}$
As $\displaystyle x \geq 0 \Rightarrow \sqrt{y+6} \geq 1 \Rightarrow y \geq-5 \Rightarrow$ Range $\displaystyle =[-5, \infty)$
As Range $\displaystyle =$ codomain $\displaystyle =[-5, \infty) \Rightarrow$ f is onto.
Given that $\displaystyle f(p)=3 \Rightarrow 4 p^{2}+4 p-5=3 \Rightarrow p=1$ as $\displaystyle p \neq-2$
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